Question

In displacement method, the distance between object and screen is $96cm$. The ratio of lengths of two images formed by a converging lens placed between them is $4$. Then

• A. Ratio of the length of object to the length of shorter image is $2$
• B. Distance between the two positions of the lens is $32cm$
• C. Focal length of the lens is $64/3cm$
• D. When the shorter image is formed on screen, distance of the lens from the screen is $32cm$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Focal length of the lens is $64/3cm$
B Ratio of the length of object to the length of shorter image is $2$
C Distance between the two positions of the lens is $32cm$
D When the shorter image is formed on screen, distance of the lens from the screen is $32cm$

$\frac{h_1}{h_2}=4$
$\frac{h_1}{h_2}\times \frac{h_2}{h_0}=1=\Rightarrow \frac{h_2}{h_0}=0$
When shorter image is formed, magnification $=2$
$\Rightarrow \left|\frac{v}{u}\right|=2$ and $|v|+|u|=96$
$\Rightarrow |v|=64$ and $|u|=32$
Hence, object distance $=32cm$ and second lens position can be $96-32=64cm$
Distance between two positions $=32cm$
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{64}-\frac{1}{(-32)}=\frac{1}{f}\Rightarrow f=\frac{64}{3}cm$.